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For a constant $0<A<\pi$, and natural $n$ I want to find the principal value of the integral:

$$\int_0^\pi \frac{\cos nt}{\cos t - \cos A} dt$$

First of all, I'm not certain what function in the complex plane I should look at. Using $$f(z) \equiv \frac{e^{inz}}{e^{iz}-\cos A}$$ or something similar doesn't work, because there is no obvious way to recover the original function from that (no simple relation of one being the imaginary part of the other etc). I tried expressing the difference of cosines as a product of sines, but I don't see how this gets me anywhere.

So: what $f(z)$ should I choose?

Then, I need to find an appropriate contour on the complex plane, and surround the singularity at $A$ with a (half?)circle of radius $\epsilon$, then take the limit $\epsilon \to 0$. I've tried a rectangle and a semicircle, but without an appropriate function to analyse, it's difficult to say what would work. In any case, for the function I tried (i.e. one with all the $t$'s replaced by $z$'s), I didn't get anywhere.

I'm not looking for an answer, just a hint on what function to consider, and on what contour.

Edit:

Following Mhenni Benghorbal's hint, I arrive at:

$$I=\frac{(-1)^{n+1}}{2i}\oint_{|z|=1} \frac{z^{2n}+1}{z^n (z+e^{iA})(z+e^{-iA})}dz$$

The problem is, the singularities are on the unit circle along which I'm integrating. I'm not sure how to deal with that.

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    $\begingroup$ Hint: $\cos$ is even, and if you write $z = e^{it}$, you have $\cos t = \frac12(z+z^{-1})$. $\endgroup$ – Daniel Fischer Nov 13 '13 at 20:16
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First, write the integral as

$$ \int_0^\pi \frac{\cos nt}{\cos t - \cos A} dt=\frac{1}{2}\int_{-\pi}^{\pi} \frac{\cos nt}{\cos t - \cos A} dt. $$

ii) Using the change of variables $ t=u-\pi $ gives

$$ \frac{(-1)^{n+1}}{2}\int_{0}^{2\pi} \frac{\cos nt}{\cos t - \cos A} dt. $$

iii) You can use residue theorem to finish the problem. Put $z=e^{iu}$ and use the identity

$$ \cos x = \frac{e^{iu}+e^{-iu}}{2}. $$

Can you finish it now? See technique.

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  • $\begingroup$ In the second bit, shouldn't the denominator be $\cos t + \cos A$? Since $t=u-\pi$, we get $\cos t = -\cos u$, and the additional minus sign that gives rise to $n+1$ in the exponent will come from cancelling $-\cos u - \cos A$? $\endgroup$ – Spine Feast Nov 13 '13 at 21:10
  • $\begingroup$ @DepeHb: In fact it does not matter. But you can write the integral in terms of $u$ sincr it is a dummy variable. As I said just double check the calculations and work out the problem step by step. $\endgroup$ – Mhenni Benghorbal Nov 13 '13 at 21:13
  • $\begingroup$ You're right, the plus wouldn't even make sense since it would remove the singularity... but I'm still getting it... $\endgroup$ – Spine Feast Nov 13 '13 at 21:15
  • $\begingroup$ Alright, I'm getting an answer, but I'm worried about how the singular integral seemingly stopped being singular after a substitution. I was meant to calculate the principal value, but here, this problem disappeared altogether. What's the reason? $\endgroup$ – Spine Feast Nov 13 '13 at 21:27
  • $\begingroup$ I updated the question with my progress. $\endgroup$ – Spine Feast Nov 13 '13 at 21:32
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The problem is, the singularities are on the unit circle along which I'm integrating. I'm not sure how to deal with that.

Right, it's a singular integral, and that means you have singularities on the contour. To deal with it, you deform the contour a little, by replacing the part near the poles on the unit circle with small (almost) semicircles in the unit disk ($\lvert z + e^{\pm iA}\rvert = \varepsilon,\, \lvert z\rvert < 1$). You can the use the residue theorem to compute the integral over the deformed contour.

The Cauchy principal value corresponds to removing small symmetric arcs around the poles, and letting the length of the removed arcs shrink to $0$.

In the deformed contour, we replaced the removed arcs with the (almost) semicircles inside the unit disk. So it remains to subtract the integrals over these small (almost) semicircles, and let their radius shrink to $0$. In the limit, that means we add half the residue in the poles on the unit circle (informally, a simple pole on the unit circle lies half inside, and half outside the unit disk).

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  • $\begingroup$ So I want to get rid of the singularities by letting my contour avoid them. But why not include them (have the semicircles stick out of my unit circle, instead of sinking in), if we're letting the radius tend to zero anyway? Or does it not matter? Shouldn't there be a difference of two residues? $\endgroup$ – Spine Feast Nov 13 '13 at 21:59
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    $\begingroup$ It doesn't matter, you could also let the semicircles bump out (then they'd be slightly more than semi). Then the residues in the poles on the unit circle would be included in the contour integral, and for the principal value, you'd subtract half the residue in these poles. End result is the same of course. $\endgroup$ – Daniel Fischer Nov 13 '13 at 22:03
  • $\begingroup$ So my answer will be that $\oint_{|z|=1} f(z)dz = 2\pi i Res(f(z),z=0) - \lim_{\epsilon \to 0}I_{\epsilon}^{(1)} - I_{\epsilon}^{(2)}$, where the $I$'s denote the integrals over small almost-semicircles? $\endgroup$ – Spine Feast Nov 13 '13 at 22:06
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    $\begingroup$ Yes, but as I said, the limits of the $I_\epsilon^{(\ast)}$ are (minus, because of orientation) $2\pi i$ times half the residue in the pole at $-e^{\pm iA}$ - I guess I left the $2\pi i$ implicit, sorry - so it's $2\pi i \operatorname{Res}(f(z),z=0) + \pi i \operatorname{Res}(f(z),z=-e^{iA}) + \pi i\operatorname{Res}(f(z),z=-e^{-iA})$. $\endgroup$ – Daniel Fischer Nov 13 '13 at 22:11

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